24.777 777 777 777 777 772 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 24.777 777 777 777 777 772 1(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
24.777 777 777 777 777 772 1(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 24.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

24(10) =

1 1000(2)


3. Convert to binary (base 2) the fractional part: 0.777 777 777 777 777 772 1.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.777 777 777 777 777 772 1 × 2 = 1 + 0.555 555 555 555 555 544 2;
  • 2) 0.555 555 555 555 555 544 2 × 2 = 1 + 0.111 111 111 111 111 088 4;
  • 3) 0.111 111 111 111 111 088 4 × 2 = 0 + 0.222 222 222 222 222 176 8;
  • 4) 0.222 222 222 222 222 176 8 × 2 = 0 + 0.444 444 444 444 444 353 6;
  • 5) 0.444 444 444 444 444 353 6 × 2 = 0 + 0.888 888 888 888 888 707 2;
  • 6) 0.888 888 888 888 888 707 2 × 2 = 1 + 0.777 777 777 777 777 414 4;
  • 7) 0.777 777 777 777 777 414 4 × 2 = 1 + 0.555 555 555 555 554 828 8;
  • 8) 0.555 555 555 555 554 828 8 × 2 = 1 + 0.111 111 111 111 109 657 6;
  • 9) 0.111 111 111 111 109 657 6 × 2 = 0 + 0.222 222 222 222 219 315 2;
  • 10) 0.222 222 222 222 219 315 2 × 2 = 0 + 0.444 444 444 444 438 630 4;
  • 11) 0.444 444 444 444 438 630 4 × 2 = 0 + 0.888 888 888 888 877 260 8;
  • 12) 0.888 888 888 888 877 260 8 × 2 = 1 + 0.777 777 777 777 754 521 6;
  • 13) 0.777 777 777 777 754 521 6 × 2 = 1 + 0.555 555 555 555 509 043 2;
  • 14) 0.555 555 555 555 509 043 2 × 2 = 1 + 0.111 111 111 111 018 086 4;
  • 15) 0.111 111 111 111 018 086 4 × 2 = 0 + 0.222 222 222 222 036 172 8;
  • 16) 0.222 222 222 222 036 172 8 × 2 = 0 + 0.444 444 444 444 072 345 6;
  • 17) 0.444 444 444 444 072 345 6 × 2 = 0 + 0.888 888 888 888 144 691 2;
  • 18) 0.888 888 888 888 144 691 2 × 2 = 1 + 0.777 777 777 776 289 382 4;
  • 19) 0.777 777 777 776 289 382 4 × 2 = 1 + 0.555 555 555 552 578 764 8;
  • 20) 0.555 555 555 552 578 764 8 × 2 = 1 + 0.111 111 111 105 157 529 6;
  • 21) 0.111 111 111 105 157 529 6 × 2 = 0 + 0.222 222 222 210 315 059 2;
  • 22) 0.222 222 222 210 315 059 2 × 2 = 0 + 0.444 444 444 420 630 118 4;
  • 23) 0.444 444 444 420 630 118 4 × 2 = 0 + 0.888 888 888 841 260 236 8;
  • 24) 0.888 888 888 841 260 236 8 × 2 = 1 + 0.777 777 777 682 520 473 6;
  • 25) 0.777 777 777 682 520 473 6 × 2 = 1 + 0.555 555 555 365 040 947 2;
  • 26) 0.555 555 555 365 040 947 2 × 2 = 1 + 0.111 111 110 730 081 894 4;
  • 27) 0.111 111 110 730 081 894 4 × 2 = 0 + 0.222 222 221 460 163 788 8;
  • 28) 0.222 222 221 460 163 788 8 × 2 = 0 + 0.444 444 442 920 327 577 6;
  • 29) 0.444 444 442 920 327 577 6 × 2 = 0 + 0.888 888 885 840 655 155 2;
  • 30) 0.888 888 885 840 655 155 2 × 2 = 1 + 0.777 777 771 681 310 310 4;
  • 31) 0.777 777 771 681 310 310 4 × 2 = 1 + 0.555 555 543 362 620 620 8;
  • 32) 0.555 555 543 362 620 620 8 × 2 = 1 + 0.111 111 086 725 241 241 6;
  • 33) 0.111 111 086 725 241 241 6 × 2 = 0 + 0.222 222 173 450 482 483 2;
  • 34) 0.222 222 173 450 482 483 2 × 2 = 0 + 0.444 444 346 900 964 966 4;
  • 35) 0.444 444 346 900 964 966 4 × 2 = 0 + 0.888 888 693 801 929 932 8;
  • 36) 0.888 888 693 801 929 932 8 × 2 = 1 + 0.777 777 387 603 859 865 6;
  • 37) 0.777 777 387 603 859 865 6 × 2 = 1 + 0.555 554 775 207 719 731 2;
  • 38) 0.555 554 775 207 719 731 2 × 2 = 1 + 0.111 109 550 415 439 462 4;
  • 39) 0.111 109 550 415 439 462 4 × 2 = 0 + 0.222 219 100 830 878 924 8;
  • 40) 0.222 219 100 830 878 924 8 × 2 = 0 + 0.444 438 201 661 757 849 6;
  • 41) 0.444 438 201 661 757 849 6 × 2 = 0 + 0.888 876 403 323 515 699 2;
  • 42) 0.888 876 403 323 515 699 2 × 2 = 1 + 0.777 752 806 647 031 398 4;
  • 43) 0.777 752 806 647 031 398 4 × 2 = 1 + 0.555 505 613 294 062 796 8;
  • 44) 0.555 505 613 294 062 796 8 × 2 = 1 + 0.111 011 226 588 125 593 6;
  • 45) 0.111 011 226 588 125 593 6 × 2 = 0 + 0.222 022 453 176 251 187 2;
  • 46) 0.222 022 453 176 251 187 2 × 2 = 0 + 0.444 044 906 352 502 374 4;
  • 47) 0.444 044 906 352 502 374 4 × 2 = 0 + 0.888 089 812 705 004 748 8;
  • 48) 0.888 089 812 705 004 748 8 × 2 = 1 + 0.776 179 625 410 009 497 6;
  • 49) 0.776 179 625 410 009 497 6 × 2 = 1 + 0.552 359 250 820 018 995 2;
  • 50) 0.552 359 250 820 018 995 2 × 2 = 1 + 0.104 718 501 640 037 990 4;
  • 51) 0.104 718 501 640 037 990 4 × 2 = 0 + 0.209 437 003 280 075 980 8;
  • 52) 0.209 437 003 280 075 980 8 × 2 = 0 + 0.418 874 006 560 151 961 6;
  • 53) 0.418 874 006 560 151 961 6 × 2 = 0 + 0.837 748 013 120 303 923 2;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.777 777 777 777 777 772 1(10) =

0.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

5. Positive number before normalization:

24.777 777 777 777 777 772 1(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


24.777 777 777 777 777 772 1(10) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) =

1 1000.1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 20 =

1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0(2) × 24


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 4

Mantissa (not normalized):
1.1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

4 + 2(11-1) - 1 =

(4 + 1 023)(10) =

1 027(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1027(10) =

100 0000 0011(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1 1000 =

1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0011

Mantissa (52 bits) =
1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001


Decimal number 24.777 777 777 777 777 772 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0011 - 1000 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100